Would the √2 be an infinitely reducible fraction if written as one?

Question

As the questions states, would the $\sqrt{2}$ be an infinitely reducible fraction if written as one?

Answer 1

Yes, this is a standard way to prove that $\sqrt{2}$ is not rational: if it were, then we can 'infinitely reduce' it as follows:

That is, assuming: $\sqrt{2} = \frac{p}{q}$ where $p$ and $q$ are whole numbers

we get: $2 = \frac{p^2}{q^2}$

and thus: $2q^2 = p^2$

and thus $p^2$ is even

and thus $p$ is even. Say $p = 2k$ for some whole number $k$

So then $2q^2 = (2k)^2 = 4k^2$

So $q^2 = 2k^2$

So $q^2$ is even

So $q$ is even. Say $q = 2m$ for some whole number $m$

And thus we can reduce:

$\sqrt{2} = \frac{p}{q} = \frac{2k}{2m} = \frac{m}{n}$

... and using the exact same reasoning, we can reduce that fraction as well! ... and thus it becomes 'infinitely reducible'.

But of course, rational numbers are not infinitely reducible like this: given that $q$ is a whole number, we can't keep dividing it by two, for that would mean that there are an infinite number of positive numbers smaller than $q$, which is impossible.

Hence, $\sqrt{2}$ cannot be a rational number.

This proof is an example of an 'infinite descent' proof, which basically combines a proof by strong induction and a proof by contradiction.